{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "

Chapter 13: D.c. circuit theory

" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 1, page no. 168

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#a) Find the unknown currents (b) Determine the value of e.m.f. E\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "Iab = 50;# in ampere\n", "Ibc = 20;# in ampere\n", "Iec = 15;# in ampere\n", "Idf = 120;# in ampere\n", "Ifg = 40;# in ampere\n", "Iab = 50;# in ampere\n", "I = 2;# in ampere\n", "V1 = 4;# in volts\n", "V2 = 3;# in volts\n", "V3 = 6;# in volts\n", "R1 = 1;# in ohms\n", "R2 = 2;# in ohms\n", "R3 = 2.5;# in ohms\n", "R4 = 1.5;# in ohms\n", "\n", "#calculation:\n", "I1 = Iab - Ibc\n", "I2 = Ibc + Iec\n", "I3 = I1 - Idf\n", "I4 = Iec - I3\n", "I5 = Idf - Ifg\n", "# Applying Kirchhoff\u2019s voltage law and moving clockwise around the loop of Figure 13.3(b) starting at point A:\n", "E = I*R2 + I*R3 + I*R4 + I*R1 - V2 - V3 + V1\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n (a) unknown currents I1, I2, I3, I4, I5 are \",I1,\"A, \", I2,\"A, \", I3,\"A, \", I4,\"A, \", I5,\"A respetively\\n\"\n", "print \"\\n (b) value of e.m.f. E = \",E,\" Volts\\n\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " (a) unknown currents I1, I2, I3, I4, I5 are 30 A, 35 A, -90 A, 105 A, 80 A respetively\n", "\n", "\n", " (b) value of e.m.f. E = 9.0 Volts\n" ] } ], "prompt_number": 1 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 2, page no. 168

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E1 = 4;# in volts\n", "E2 = 2;# in volts\n", "r1 = 2;# in ohms\n", "r2 = 1;# in ohms\n", "R = 4;# in ohms\n", "\n", "#calculation:\n", "#E1 = I1*(r1 + R) + I2*R,\n", "#E2 = I1*R + (R + r2)*I2,\n", "I2 = (E1*R - E2*(r1 + R))/(R**2 - (R+r1)*(R + r2))\n", "I1 = (E1 - I2*R)/(r1 + R)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n currents flowing are, I1 = \",round(I1,3),\"A, and I2 = \",round(-1*I2,3),\"A, \"\n", "print \"and current flowing in middle branch is I1 - I2 = \", round(I1 + I2, 3),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " currents flowing are, I1 = 0.857 A, and I2 = 0.286 A, and current flowing in middle branch is I1 - I2 = 0.571 A" ] } ], "prompt_number": 2 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 3, page no. 169

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E1 = 4;# in volts\n", "E2 = 12;# in volts\n", "R1 = 0.5;# in ohms\n", "R2 = 2;# in ohms\n", "R3 = 5;# in ohms\n", "\n", "#calculation:\n", "#E1 + E2 = I1*R1 + I2*R2\n", "#E2 = - I1*R3 + I2*(R2 + R3) \n", "I2 = ((E1 + E2)*R3 + E2*R1)/(R2*R3 + (R2+R3)*R1)\n", "I1 = (E1 + E2 - I2*R2)/R1\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n currents flowing are, I1 = \",round(I1,2),\"A, and I2 = \",round(I2,2),\"A\"\n", "print \" and in R3 branch is I1 - I2\", round(I1 - I2,2),\" A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " currents flowing are, I1 = 6.52 A, and I2 = 6.37 A and in R3 branch is I1 - I2 0.15 A" ] } ], "prompt_number": 3 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 4, page no. 170

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#determine the currents in each of the resistors\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "V = 54;# in volts\n", "I = 8;# in Amps\n", "R1 = 2;# in ohms\n", "R2 = 11;# in ohms\n", "R3 = 14;# in ohms\n", "R4 = 3;# in ohms\n", "R5 = 32;# in ohms\n", "\n", "#calculation:\n", "#V = (R1 + R2)*I1 - R2*I2\n", "#0 = (R1 + R3)*I1 - R5*I2 - R3*I\n", "I1 = V*R5/((R1 + R2)*R5 - (R1 + R3)*R2 + R3*I)\n", "I2 = -1*(V - I1*(R2 + R1))/R2\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 2 ohm resistor = \",round(I1,0),\"A,\"\n", "print \"\\n the current flowing in the 14 ohm resistor = \",round(I - I1,0),\"A,\"\n", "print \"\\n the current flowing in the 32 ohm resistor = \",round(I2,0),\"A,\"\n", "print \"\\n the current flowing in the 11 ohm resistor = \",round(I1 - I2,0),\"A,\"\n", "print \"\\n the current flowing in the 3 ohm resistor = \",round(I - I1 + I2,0),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 2 ohm resistor = 5.0 A,\n", "\n", " the current flowing in the 14 ohm resistor = 3.0 A,\n", "\n", " the current flowing in the 32 ohm resistor = 1.0 A,\n", "\n", " the current flowing in the 11 ohm resistor = 4.0 A,\n", "\n", " the current flowing in the 3 ohm resistor = 4.0 A" ] } ], "prompt_number": 4 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 5, page no. 171

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Determine the current in each branch of the network by using the superposition theorem\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E1 = 4;# in volts\n", "E2 = 2;# in Volts\n", "R1 = 2;# in ohms\n", "R2 = 1;# in ohms\n", "R = 4;# in ohms\n", "\n", "#calculation:\n", "I1 = E1/(R1 + (R*R2/(R + R2)))\n", "I2 = (R2/(R + R2))*I1\n", "I3 = (R/(R + R2))*I1\n", "I4 = E2/(R2 + (R*R1/(R + R1)))\n", "I5 = (R1/(R + R1))*I4\n", "I6 = (R/(R + R1))*I4\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n Resultant current flowing through source 1 = \",round(I1 - I6,3),\"A,\"\n", "print \"\\n Resultant current flowing through source 2 = \",round(I4 - I3,3),\"A,\"\n", "print \"\\n Resultant current flowing through resistor R, = \",round(I2 + I5,3),\"A,\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " Resultant current flowing through source 1 = 0.857 A,\n", "\n", " Resultant current flowing through source 2 = -0.286 A,\n", "\n", " Resultant current flowing through resistor R, = 0.571 A," ] } ], "prompt_number": 5 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 6, page no. 173

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#find, using the superposition theorem, (a) the current flowing in and the pd across the 18 ohm resistor, \n", "#(b) the current in the 8 V battery and (c) the current in the 3 V battery.\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E1 = 8;# in volts\n", "E2 = 3;# in Volts\n", "R1 = 3;# in ohms\n", "R2 = 2;# in ohms\n", "R = 18;# in ohms\n", "\n", "#calculation:\n", "I1 = E1/(R1 + (R*R2/(R + R2)))\n", "I2 = (R/(R + R2))*I1\n", "I3 = (R2/(R + R2))*I1\n", "I4 = E2/(R2 + (R*R1/(R + R1)))\n", "I5 = (R/(R + R1))*I4\n", "I6 = (R1/(R + R1))*I4\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n (a)Resultant current in the 18 ohm resistor= \",round(I3 - I6,3),\"A \"\n", "print \"and P.d. across the 18 ohm resistor\",round((I3-I6)*R,3),\"V\"\n", "print \"\\n (b)the current in the 8 V battery= \",round(I1 + I5,3),\"A\"\n", "print \"\\n (c)current in the 3 V battery = \",round(I2 + I4,3),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " (a)Resultant current in the 18 ohm resistor= 0.073 A and P.d. across the 18 ohm resistor 1.313 V\n", "\n", " (b)the current in the 8 V battery= 2.229 A\n", "\n", " (c)current in the 3 V battery = 2.156 A" ] } ], "prompt_number": 6 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 7, page no. 176

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Use Th\u00b4evenin\u2019s theorem to find the current flowing in the 10 ohm resistor for the circuit\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "V = 10;# in volts\n", "R1 = 2;# in ohms\n", "R2 = 8;# in ohms\n", "R3 = 5;# in ohms\n", "R = 10;# in ohms\n", "\n", "#calculation:\n", "I1 = V/(R1 + R2)\n", "E = I1*R2\n", "r = R3 + R1*R2/(R1 + R2)\n", "I = E/(R + r)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 10 ohm resistor = 0.482 A" ] } ], "prompt_number": 7 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 8, page no. 177

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#determine the current in the 0.8 ohm resistor using Th\u00b4evenin\u2019s theorem.\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "V = 12;# in volts\n", "R1 = 1;# in ohms\n", "R2 = 4;# in ohms\n", "R3 = 5;# in ohms\n", "R = 0.8;# in ohms\n", "\n", "#calculation:\n", "I1 = V/(R1 + R2 + R3)\n", "E = I1*R2\n", "r = R2*(R1 + R3)/(R1 + R2 + R3)\n", "I = E/(R + r)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 0.8 ohm resistor = \",round(I,3),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 0.8 ohm resistor = 1.5 A" ] } ], "prompt_number": 8 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 9, page no. 178

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#determine the current in the 4 ohm resistor using Th\u00b4evenin\u2019s theorem.Find also the power dissipated in the 4 ohm resistor.\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E1 = 4;# in volts\n", "E2 = 2;# in volts\n", "R1 = 2;# in ohms\n", "R2 = 1;# in ohms\n", "R = 4;# in ohms\n", "\n", "#calculation:\n", "I1 = (E1 - E2)/(R1 + R2)\n", "E = E1 - I1*R1\n", "r = R2*R1/(R1 + R2)\n", "I = E/(R + r)\n", "P = R*I**2\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A and power dissipated in the 4 ohm resistor = \",round(P,3),\"W\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 4 ohm resistor = 0.571 A and power dissipated in the 4 ohm resistor = 1.306 W" ] } ], "prompt_number": 9 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 10, page no. 178

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#determine the current in the 3 ohm resistor using Th\u00b4evenin\u2019s theorem\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "V = 24;# in volts\n", "R = 3;# in ohms\n", "R1 = 20;# in ohms\n", "R2 = 5;# in ohms\n", "R3 = 10;# in ohms\n", "R4 = 5/3;# in ohms\n", "\n", "#calculation:\n", "E = R3*V/(R3 + R2)\n", "r = R4 + R3*R2/(R3 + R2)\n", "I = E/(R + r)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 3 ohm resistor = 2.0 A" ] } ], "prompt_number": 10 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 11, page no. 179

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#determine the current in the 32 ohm resistor using Th\u00b4evenin\u2019s theorem\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E = 54;# in volts\n", "R1 = 2;# in ohms\n", "R2 = 14;# in ohms\n", "R3 = 3;# in ohms\n", "R4 = 11;# in ohms\n", "R5 = 32;# in ohms\n", "\n", "#calculation:\n", "Vac = R1*E/(R1 + R4)\n", "Vbc = R2*E/(R2 + R3)\n", "V = Vbc - Vac\n", "r = R4*R1/(R1 + R4) + R3*R2/(R3 + R2)\n", "I = V/(R5 + r)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 32 ohm resistor = \",round(I,0),\"A flowing from A to B\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 32 ohm resistor = 1.0 A flowing from A to B" ] } ], "prompt_number": 11 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 12, page no. 181

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#determine the current in the 10 ohm resistor using Norton's theorem\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E = 10;# in volts\n", "R1 = 2;# in ohms\n", "R2 = 8;# in ohms\n", "R3 = 5;# in ohms\n", "R = 10;# in ohms\n", "\n", "#calculation:\n", "Isc = E/R1\n", "r = R1*R2/(R1 + R2)\n", "I = r*Isc/(r + R3 + R)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 10 ohm resistor = 0.482 A" ] } ], "prompt_number": 12 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 13, page no. 182

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#determine the current in the 4 ohm resistor using Norton's theorem\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E1 = 4;# in volts\n", "E2 = 2;# in volts\n", "R1 = 2;# in ohms\n", "R2 = 1;# in ohms\n", "R = 4;# in ohms\n", "\n", "#calculation:\n", "Isc = E1/R1 + E2/R2\n", "r = R1*R2/(R1 + R2)\n", "I = r*Isc/(r + R)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 4 ohm resistor = 0.571 A" ] } ], "prompt_number": 13 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 14, page no. 182

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#determine the current in the 3 ohm resistor using Norton's theorem\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "V = 24;# in volts\n", "R = 3;# in ohms\n", "R1 = 20;# in ohms\n", "R2 = 5;# in ohms\n", "R3 = 10;# in ohms\n", "R4 = 5/3;# in ohms\n", "\n", "#calculation:\n", "Isc = V/R2\n", "r = R3*R2/(R3 + R2)\n", "I = r*Isc/(r + R4 + R)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 3 ohm resistor = 2.0 A" ] } ], "prompt_number": 14 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 15, page no. 183

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#determine the current in the 2 ohm resistor.\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "I = 15;# in Amps\n", "R = 2;# in ohms\n", "R1 = 6;# in ohms\n", "R2 = 4;# in ohms\n", "R3 = 8;# in ohms\n", "R4 = 7;# in ohms\n", "\n", "#calculation:\n", "Isc = R1*I/(R1 + R2)\n", "r = (R1 + R2)*(R3 + R4)/(R3 + R1 + R4 + R2)\n", "I = r*Isc/(r + R)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n the current flowing in the 2 ohm resistor = \",round(I,2),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " the current flowing in the 2 ohm resistor = 6.75 A" ] } ], "prompt_number": 15 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 16, page no. 185

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Convert the circuit to an equivalent Norton network.\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "V = 10;# in Volts\n", "R = 2;# in ohms\n", "\n", "#calculation:\n", "Isc = V/R\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n equivalent Norton network contains Current Source of amp = \",round(Isc,0),\"A and a resistor of \",R,\" ohm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " equivalent Norton network contains Current Source of amp = 5.0 A and a resistor of 2 ohm" ] } ], "prompt_number": 16 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 17, page no. 185

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Convert the circuit to an equivalent Thevenin network.\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "Isc = 4;# in Amps\n", "R = 3;# in ohms\n", "\n", "#calculation:\n", "E = Isc*R\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,0),\"V and a resistor of \",R,\" ohm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " equivalent Thevenin network contains Voltage Source of 12.0 V and a resistor of 3 ohm" ] } ], "prompt_number": 17 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 18, page no. 185

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#(a)Convert the circuit to an equivalent Thevenin networkby initially converting to a Norton equivalent circuit.\n", "#(b)Determine the current flowing in the 1.8 ohm resistor.\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E1 = 12;# in Volts\n", "E2 = 24;# in Volts\n", "R1 = 3;# in ohms\n", "R2 = 2;# in ohms\n", "R = 1.8;# in ohms\n", "\n", "#calculation:\n", "Isc1 = E1/R1\n", "Isc2 = E2/R2\n", "I1 = Isc1 + Isc2\n", "r = R1*R2/(R1 + R2)\n", "E = I1*r\n", "I = E/(r + R)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n (A)equivalent Norton network contains Current Source of \",round(I1,1),\"A and a resistor of \",r,\" ohm\"\n", "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",r,\" ohm\"\n", "print \"\\n (B)the current flowing in the 1.8 ohm resistor is \",round(I,1),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " (A)equivalent Norton network contains Current Source of 16.0 A and a resistor of 1.2 ohm\n", "\n", " equivalent Thevenin network contains Voltage Source of 19.2 V and a resistor of 1.2 ohm\n", "\n", " (B)the current flowing in the 1.8 ohm resistor is 6.4 A" ] } ], "prompt_number": 18 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 19, page no. 186

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#(a)Determine by successive conversions between Th\u00b4evenin and Norton equivalent networks \n", "#a Thevenin equivalent circuit for terminals AB.\n", "#(b)determine the current flowing in the 200 ohm resistance.\n", "from __future__ import division\n", "import math\n", "#initializing the variables:\n", "E1 = 10;# in Volts\n", "E2 = 6;# in Volts\n", "I1 = 0.001;#in Amp\n", "R1 = 2000;# in ohms\n", "R2 = 3000;# in ohms\n", "R3 = 600;# in ohms\n", "R = 200;# in ohms\n", "\n", "#calculation:\n", "Isc1 = E1/R1\n", "Isc2 = E2/R2\n", "I2 = Isc1 + Isc2\n", "r1 = R1*R2/(R1 + R2)\n", "E = I2*r1 - I1*R3\n", "r = r1 + R3\n", "I = E/(r + R)\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n (A)equivalent Norton network contains Current Source of \",round(I2*1000,0),\"mA and a resistor of \",round(r1/1000,1),\" Kohm\"\n", "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",round(r1/1000,1),\" Kohm\"\n", "print \"\\n (B)the current flowing in the 200 ohm resistor is \",round(I*1000,1),\"mA\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " (A)equivalent Norton network contains Current Source of 7.0 mA and a resistor of 1.2 Kohm\n", "\n", " equivalent Thevenin network contains Voltage Source of 7.8 V and a resistor of 1.2 Kohm\n", "\n", " (B)the current flowing in the 200 ohm resistor is 3.9 mA" ] } ], "prompt_number": 19 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 20, page no. 188

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the power dissipated by the load in each case.\n", "#Plot a graph of RL (horizontally) against power (vertically) and determine the maximum power dissipated.\n", "from __future__ import division\n", "import math\n", "#from pylab import *\n", "%pylab inline\n", "#initializing the variables:\n", "E = 6;# in Volts\n", "R = 2.5;# in ohms\n", "\n", "#calculation:\n", "RL = []\n", "P = []\n", "k = []\n", "for h in range(6):\n", " RL.append(h - 0.5)\n", " k = h - 0.5\n", " P.append(k*((E/(R + k))**2))\n", "fig = plt.figure()\n", "ax = fig.add_subplot(1, 1, 1)\n", "ax.plot(RL,P,'-')\n", "#plot(RL,P,'-')\n", "xlabel('RL(ohm)')\n", "ylabel('Power(W)')\n", "show()\n", "Pmax= R*(E/(2*R))**2\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \"\\n A graph of RL against P is shown in Figure\" \n", "print \" The maximum value of power is\", Pmax,\"W which occurs when RL =\",R,\"ohm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n", "For more information, type 'help(pylab)'." ] }, { "output_type": "display_data", "png": "iVBORw0KGgoAAAANSUhEUgAAAX8AAAEMCAYAAAAs8rYIAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz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\nZLqKDYkAAAAASUVORK5CYII=\n" }, { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", "\n", " A graph of RL against P is shown in Figure\n", " The maximum value of power is 3.6 W which occurs when RL = 2.5 ohm" ] } ], "prompt_number": 1 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 21, page no. 188

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#State the value of load resistance that gives maximum power dissipation and determine the value of this power.\n", "from __future__ import division\n", "import math\n", "from pylab import *\n", "#initializing the variables:\n", "E = 30;# in Volts\n", "R = 1.5;# in ohms\n", "\n", "#calculation:\n", "RL = R\n", "I = E/(R + RL)\n", "P = I**2*RL\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \" The maximum value of power is\", P,\"W which occurs when RL =\",RL,\"ohm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", " The maximum value of power is 150.0 W which occurs when RL = 1.5 ohm" ] } ], "prompt_number": 21 }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 22, page no. 189

" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Find the value of the load resistor RL that gives maximum power dissipation and determine the value of this power.\n", "from __future__ import division\n", "import math\n", "from pylab import *\n", "#initializing the variables:\n", "V = 15;# in Volts\n", "R1 = 3;# in ohms\n", "R2 = 12;# in ohms\n", "\n", "#calculation:\n", "E = (R2/(R2+ R1))*V\n", "r = R1*R2/(R1 + R2)\n", "RL = r\n", "I = E/(r + RL)\n", "P = I**2*RL\n", "\n", "#Results\n", "print \"\\n\\n Result \\n\\n\"\n", "print \" The maximum value of power is\", P,\"W which occurs when Total Load RL =\",RL,\"ohm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "\n", " Result \n", "\n", "\n", " The maximum value of power is 15.0 W which occurs when Total Load RL = 2.4 ohm" ] } ], "prompt_number": 22 } ], "metadata": {} } ] }